foci y=6a^2+5a+10-(4a^2+6a+12)

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`x`²	`x`^□	log_□	√☐	^□√☐	≤	≥	□□	·	÷	`x`^◦	π
(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

☐²

x^☐

√☐

^□√☐

□□

log_□

θ

∞

∫

ddx

≥	≤	·	÷	`x`^◦	(☐)	\|☐\|	(`f` ◦ `g`)	`f`(`x`)	ln	`e`^☐
(☐)^′	∂∂`x`	∫_□^□	lim	∑	sin	cos	tan	cot	csc	sec

simplify solve for expand factor rationalize

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vértices y=(x+5)²

Solution

Mínimo (−5, 0)

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Solve by:

Encontre o vértice usando o ponto médio das raízes

Encontrar o vértice usando a forma polinomial

Encontrar o vértice usando a forma da parábola

Encontrar o vértice usando a forma do vértice

One step at a time

y=(x+5)²

Equação de parábola na forma fatorada

O vértice de uma parábola com concavidade para cima ou para baixo na forma y=a(x−m)(x−n) é a média das raízes x_v=m+n2

y=(x+5)²

Os parâmetros da parábola são:

a=1, m=−5, n=−5

x_v=m+n2

x_v=(−5)+(−5)2

Simplificar

x_v=−5

Insira x_v=−5 para encontrar o valor y_v

y_v=0

Portanto, o vértice da parábola é

(−5, 0)

Se a<0, então o vértice é um valor máximo
Si a>0, então o vértice é um valor mínimo
a=1

Mínimo (−5, 0)

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−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
▭ \|▭	)	.	0	=	+	`y`

foci y=6a^2+5a+10-(4a^2+6a+12)

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foci y=6a^2+5a+10-(4a^2+6a+12)

Solution

Solution steps

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